Method 1
Expand the right-hand side and collect the $b x$ terms

$\begin{array}{rcl} {(x + b)}^{2} & = & (x + b) (x + b) \\ = & x^{2} + b x + b x + b^{2} \\ = & x^{2} + 2 b x + b^{2} \end{array}$

Make $x^{2} - 12 x + a$ equal to $x^{2} + 2 b x + b^{2}$ by comparing each term
The $x^{2}$ terms are equal
The $x$ terms are equal if

$- 12 = 2 b$

Solve this equation to find $b$ (by dividing both sides by 2)

$\begin{array}{rcl} - \frac{12}{2} & = & b \\ - 6 & = & b \end{array}$

$b = - 6$ [1]

The number terms on the end are equal if

$a = b^{2}$

[1]

Substitute in $b = - 6$

$a = {(- 6)}^{2} a = 36$

$a = 36$ [1]

Method 2
Complete the square on $x^{2} - 12 x + a$ by replacing $x^{2} - 12 x$ with ${(x - 6)}^{2} - 36$

$x^{2} - 12 x + a = {(x - 6)}^{2} - 36 + a$

${(x - 6)}^{2} - 36 + a$ must equal ${(x + b)}^{2}$
The brackets are equal if

$b = - 6$

$b = - 6$ [1]

The numbers outside the brackets are equal if

$- 36 + a = 0$

[1]

Solve the equation to find $a$ (by adding $36$ to both sides)

$a = 36$

$a = 36$ [1]

Completing the Square (Edexcel IGCSE Maths)

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